Question

A triangle has corners at `(2 , 2 )`, ( 5, 6 )`, and`( 1, 4 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?

Answer 1

Each of the 3 segments have one endpoint at the midpoint of the given points and the other endpoint at `(2.5,4.75), (3.5,4) or (53/22,28/11)`, and lengths equal to `1.25, .5*sqrt(5) or 5*sqrt(5)/11`.

Explanation:

Repeating the points
`A(2,2), B(5,6), C(1,4)`

Midpoints
`M_(AB) (3.5,4)`, `M_(BC) (3,5)`, `M_(CA) (1.5,3)`

Slopes of segments (`k=(Delta y)/(Delta x)`, `p=-1/k`)
`AB -> k_1=(6-2)/(5-2)=4/3 -> p_1=-3/4`
`BC -> k_2=(4-6)/(1-5)=(-2)/-4=1/2 -> p_2=-2`
`CA -> k_3=(4-2)/(1-2)=4/(-1)=-2 -> p_3=-1/2`

Now to simplify the work that remains to do, be noted that the line perpendicular to a side meets one of the other sides (or both at the same time when it intercepts a vertex of the triangle). When it meets just one of the two other sides, which side is this? Answer: the LONG side (although that is intuitive it can be proved). Then we need to know the lengths of the sides of the triangle.
`AB=sqrt((5-2)^2+(6-2)^2)=sqrt (9+16)=5`
`BC=sqrt((1-5)^2+(4-5)^2)=sqrt(16+4)=2*sqrt(5)~=4.472`
`CA=sqrt((1-2)^2+(4-2)^2)=sqrt(1+4)=sqrt(5)~=2.236`
=> `AB>BC>CA`

So
line 1 perpendicular to AB meets side BC
line 2 perpendicular to BC meets side AB
line 3 perpendicular to AC meets side AB

We need the equations of the lines in which the sides AB and BC lay and the equations of the 3 perpendicular lines

Equation of the line that supports side:
`AB -> (y-2)=(4/3)(x-2)` => `y=(4x-8)/3+2` => `y=(4x-2)/3` [a]
`BC->y-4=(1/2)(x-1)`=>`y=(x-1)/2+4`=>`y=(x+7)/2`[b]

Equation of the line (passing through midpoint) perpendicular to side:
`AB -> (y-4)=(-3/4)(x-3.5)` => `y=(-3x+10.5)/4+4` => `y=(-3x+26.5)/4` [1]
`BC -> (y-5)=-2(x-3)` => `y=-2x+6+5` => `y=-2x+11` [2]
`AC -> (y-3)=-(1/2)(x-1.5)` => `y=(-x+1.5)/2+3` => `y=(-x+7.5)/2` [3]

Finding the interceptions on sides AB and BC

Combining equations [b] and [1]

`{y=(x+7)/2`
`{y=(-3x+26.5)/4` => `(x+7)/2=(-3x+26.5)/4` => `2x+14=-3x+26.5` => `5x=12.5` => `x=2.5`
`-> y=(2.5+7)/2` => `y=4.75`

We've found `R(2.5,4.75)`
The distance between `M_(AB)` and R is
`d1=sqrt((2.5-3.5)^2+(4.75-4)^2)=sqrt(1+.5625)=1.25`

Combining equations [a] and [2]

`{y=(4x-2)/3`
`{y=-2x+11` => `(4x-2)/3=-2x+11` => `4x-2=-6x+33` => `10x=35` => `x=3.5`
`-> y=-2*3.5+11` => `y=4`

We've found `S(3.5,4)`
We can find the distance between `M_(BC)` and `S`:
`d2=sqrt((3.5-3)^2+(4-5)^2)=sqrt(1.25)=5*sqrt(5)`

Combining the equations [a] and [3]

`{y=(4x-2)/3`
`{y=(-x+7.5)/2` => `(4x-2)/3=(-x+7.5)/2` => `8x-4=-3x+22.5` => `11x=26.5` => `x=53/22`
`-> y=(-53/22+15/2)/2=(-53+165)/44` => `y=28/11`

We've found `T(53/22, 28/11)`
The distance between `M_(CA)` and T is
`d3=sqrt((53/22-3/2)^2+(28/11-3)^2)=sqrt((10/11)^2+(5/11)^2)=sqrt(125)/11=5*sqrt(5)/11`